91Ó°ÊÓ

Equation manipulation is a crucial skill in mathematics, allowing us to rearrange equations to solve for unknown variables. In this exercise, we start with two parametric equations:

• \( x(t) = 3t - 1 \)
• \( y(t) = 5 - 2t \)

The goal is to express these equations in terms of \( x \) and \( y \) by eliminating the parameter \( t \).
First, we manipulate the first equation into a form that isolates \( t \). This involves reversing operations applied to \( t \). For instance, since \( t \) is multiplied by 3 and then reduced by 1, we reverse these operations:

• Add 1 to both sides to undo the subtraction: \( x + 1 = 3t \)
• Divide both sides by 3 to isolate \( t \): \( t = \frac{x + 1}{3} \)

This type of manipulation is fundamental in solving equations and transforming expressions.

Algebraic substitution is a method used to replace variables with expressions to simplify equations or solve problems. After isolating \( t \) from the equation \( x(t) = 3t - 1 \), we know \( t = \frac{x + 1}{3} \).
Now, we use this expression for \( t \) to substitute it into the second parametric equation for \( y(t) \), which is \( y(t) = 5 - 2t \).
By substituting, we get:

• Replace \( t \) with \( \frac{x + 1}{3} \): \( y = 5 - 2\left(\frac{x + 1}{3}\right) \)
• Simplify the multiplication: \( y = 5 - \frac{2(x + 1)}{3} \)

Using substitution in this way replaces a variable, letting us eventually express the relationship entirely in terms of \( x \) and \( y \), which is the goal of parametric elimination.

Coordinate geometry allows us to represent geometric shapes and curves using algebraic equations. In this exercise, we aim to express the curve, initially defined in parametric form, as a function in a coordinate plane.
Once the substitution is complete, we have a linear equation in standard form:

• \( y = 5 - \frac{2(x + 1)}{3} \)
• Simplify the expression: \( y = 5 - \frac{2}{3}x - \frac{2}{3} \)
• Combine like terms: \( y = \frac{16}{3} - \frac{2}{3}x \)

Now, this equation can be interpreted geometrically.
This linear equation represents a straight line on the coordinate plane. The slope of the line, \( -\frac{2}{3} \), describes how steep the line is, and the y-intercept \( \frac{16}{3} \) indicates where the line crosses the y-axis.
Such transformations help in understanding how algebraic expressions relate to geometric representations, enabling a deeper comprehension of the subject.